Count Iblis said:Fleem, it is a fraction.
Hint: split it in partial fractions.
eok20 said:The Attempt at a Solution
For |x| < 1 I can write f as a power series (since 1/(1+x^2) = sum_n (-1)^n x^(2n)) but this won't work at 1. I tried writing out the first few derivatives at 1 explicitly but things got pretty messy. Any ideas?
The 100th derivative of f(x) = x/(1+x^2) at 1 is equal to 0.
To find the 100th derivative of f(x) = x/(1+x^2) at 1, we can use the formula for the nth derivative of the quotient of two functions: (f/g)^n = (f^n - nf^(n-1)g + (n choose 2)f^(n-2)g^2 - ... - g^n)/(g^(n+1)).In this case, we have f(x) = x and g(x) = 1+x^2. Substituting these values into the formula gives us 0 as the 100th derivative at 1.
Yes, there is a simpler way to find the 100th derivative of f(x) = x/(1+x^2) at 1. We can use the fact that the function is a rational function, which means that all of its derivatives will also be rational functions. Therefore, the 100th derivative at 1 will also be a rational function with a constant value of 0.
The 100th derivative of f(x) = x/(1+x^2) at 1 represents the rate of change of the function at x = 1, after taking 100 derivatives. Since the function is a rational function, its 100th derivative at 1 is 0, indicating that the rate of change is constant at this point.
No, the 100th derivative of f(x) = x/(1+x^2) at 1 cannot be negative. As mentioned before, the function is a rational function and all of its derivatives will also be rational functions, which cannot have negative values. Therefore, the 100th derivative at 1 will always be 0.